Let -pi/6<theta<-pi/12. Suppose a1 and b1 are the roots of the equation x2 - 2xsec(theta) + 1=0 and a2 and b2 are the roots of the equation x2 + 2xtan(theta) - 1=0. If a1>b1 and a2>b2, then a1 + b2 is???

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Please find below the solution to the asked query:

x2-2xsecθ+1=0Using Shridharacharya's formula we get:x=--2secθ±-2secθ2-4×1×12×1=2secθ±4sec2θ-42=2secθ±2sec2θ-12=secθ±tan2θ=secθ±tanθAs -π6<θ<-π12, hence tanθ will be negative.tanθ=-tanθx=secθtanθa1=secθ-tanθa2=secθ+tanθx2+2xtanθ-1=0x=-2tanθ±-2tanθ2-4×1×1-12×1=-2tanθ±4tan2θ+42=-2tanθ±2tan2θ+12=-2tanθ±2sec2θ2=-tanθ±secθAs secθ is positive when -π6<θ<-π12secθ=secθx=-tanθ±secθ, hence roots arex=secθ-tanθ and x=-secθ-tanθAs a1=secθ-tanθa2=secθ+tanθNote that secθ-tanθ is positiveand a1>b1 and a2>b2b1=-secθ-tanθandb2=secθ-tanθBut in that caseb2>a2assecθ-tanθ>-secθ-tanθNo combination fits a1>b1 and a2>b2You can also see that in the graph given below.

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